The number of distinct real roots of the equation $|x+1||x+3|-4|x+2|+5=0$, is
The number of distinct real roots of the equation $|x+1||x+3|-4|x+2|+5=0$, is
Solution
$|x+1||x+3|-4|x+2|+5=0$
case-1
$\begin{aligned}
& x \leq-3 \\
& (x+1)(x+3)+4(x+2)+5=0 \\
& x^2+4 x+3+4 x+8+5=0 \\
& x^2+8 x+16=0 \\
& (x+4)^2=0 \\
& x=-4
\end{aligned}$
case-2
$\begin{aligned}
& -3 \leq x \leq-2 \\
& -x^2-4 x-3+4 x+8+5=0 \\
& -x^2+10=0 \\
& x= \pm \sqrt{10}
\end{aligned}$
case-3
$\begin{aligned}
& -2 \leq x \leq-1 \\
& -x^2-4 x-3-4 x-8+5=0 \\
& -x^2-8 x-6=0 \\
& x^2+8 x+6=0 \\
& x=\frac{-8 \pm 2 \sqrt{10}}{2}=-4 \pm \sqrt{10}
\end{aligned}$
case-4
$\begin{aligned}
& x \geq-1 \\
& x^2+4 x+3-4 x-8+5=0 \\
& x^2=0 \\
& x=0
\end{aligned}$
No. of solution $=2$
Asked in: JEE Main 2024 (08 Apr Shift 2)
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